The single most important skill in this chapter is not memorising formulas — it is asking one question the moment you read a problem: does order matter here?
Look at this scenario. Ten students are competing for a quiz team. If you are picking 3 students to form a team, the group {Ravi, Priya, Arjun} and the group {Arjun, Ravi, Priya} are the same team. Order does not matter. This is a combination (चुनाव).
Now change the scenario. You are picking 3 students to win Gold, Silver, and Bronze medals. Now Ravi-Gold / Priya-Silver / Arjun-Bronze is completely different from Arjun-Gold / Ravi-Silver / Priya-Bronze. Order matters. This is a permutation (क्रमचय).
Think of it this way:
The analogy that sticks: imagine you have 4 coloured balls — Red, Blue, Green, Yellow. If you pick 2 to place in Box-1 and Box-2 separately, that is a permutation (Red in Box-1 is different from Blue in Box-1). If you pick 2 to put in a single bag together, that is a combination (Red+Blue in the bag equals Blue+Red in the bag).
The foundational counting principle under both: the Fundamental Principle of Counting (गुणन नियम). If task A can be done in m ways and task B independently in n ways, both together can be done in m × n ways. Every PYQ in this chapter traces back to this principle.
n! = n × (n-1) × (n-2) × ... × 2 × 1
Memorise these cold:
| n | n! | |---|---| | 0 | 1 | | 1 | 1 | | 2 | 2 | | 3 | 6 | | 4 | 24 | | 5 | 120 | | 6 | 720 | | 7 | 5040 |
0! = 1 is a definition, not a coincidence. If you forget it, every combination formula breaks.
nPr = n! / (n-r)!
In plain language: you have n distinct items, you want to arrange r of them in a row. Fill the first slot in n ways, the next in (n-1) ways, and so on for r slots.
nPr = n × (n-1) × (n-2) × ... × (n-r+1) ← this is just r consecutive terms starting from n.
Example: P(6,3) = 6 × 5 × 4 = 120. Three terms. Start at 6, multiply downward, stop after 3 terms.
No need to compute 6!/3! — multiply downward for r steps.
nCr = n! / (r! × (n-r)!)
The intuition: start with all nPr arrangements, then divide by r! because within your chosen group of r items, the r! different orderings all represent the same selection.
nCr = nPr / r!
Example: C(9,3) = (9 × 8 × 7) / (3 × 2 × 1) = 504 / 6 = 84
Numerator: r terms going down from n. Denominator: r!.
Key properties:
nCr = nC(n-r) — selecting 3 from 9 is the same count as rejecting 6 from 9.nC0 = nCn = 1nC1 = nMany UP Police PYQs use this structure:
Restriction: two specific people must NOT be together.
Valid arrangements = Total arrangements - Arrangements where both are together
Step 1: Total = n!
Step 2: Treat the two restricted people as one unit. You now arrange (n-1) units, and the two people inside the unit can swap, giving (n-1)! × 2!.
Step 3: Subtract.
This subtraction approach (complementary counting) appears repeatedly. Internalise the three-step structure.
When every person in a group sends something to every other person, the count is:
n × (n-1)
Because each of the n people sends to (n-1) others. This is just nP2 = n!/(n-2)!.
Note: this is NOT nC2. Sending a card from A to B is different from B sending to A. Order matters here. Use permutation, not combination.
When your pool is built from two different categories (e.g., digits + letters), count the total pool size first, then apply nPr or nCr as the problem demands.
Pool of 10 digits (0–9) + 6 letters (A–F) = 16 characters total.
For a 4-character password with no repetition: 16 × 15 × 14 × 13 = 43680.
This is just P(16,4) computed by multiplying downward 4 steps.
For nPr, never write out full factorials. Just multiply r consecutive integers starting from n going downward.
P(6,3): Start at 6, write 3 terms → 6 × 5 × 4 = 120. P(16,4): Start at 16, write 4 terms → 16 × 15 × 14 × 13 = 43680.
Standard method (expanding factorials): ~45 seconds. This method: ~12 seconds. The difference compounds across a full exam.
For nCr, write the same r terms in the numerator as above, then divide by r!.
C(9,3): Numerator = 9 × 8 × 7 = 504. Denominator = 3! = 6. Answer = 84. C(12,4): Numerator = 12 × 11 × 10 × 9 = 11880. Denominator = 4! = 24. Answer = 495.
You never need to expand n! fully. 4 steps of multiplication vs. writing a 9-digit factorial. Saves 30+ seconds per question.
When r > n/2, swap to the smaller complement before computing.
C(9,6): Computing directly needs 6 terms in numerator. Instead, C(9,6) = C(9,3). Now only 3 terms: 9 × 8 × 7 / 6 = 84.
Rule: if r is more than half of n, subtract. Cuts your multiplication steps by half. Saves ~20 seconds on large-r combinations.
When two people must be together: glue them into one unit. n people become (n-1) units. Arrangements = (n-1)! × 2.
Not-together = Total - Together = n! - (n-1)! × 2.
For n=6: 720 - (120 × 2) = 720 - 240 = 480. Takes 10 seconds once the formula structure is memorised vs. 60+ seconds if you try to enumerate cases.
Three-word test: does the problem mention rank / position / password / arrange / row / queue / award? → Permutation.
Does it mention team / committee / group / select / choose / committee / hand? → Combination.
When you see "send cards to each other" — note that A→B and B→A are different acts. That is P, not C: use n(n-1).
Applying this before touching numbers eliminates wrong-formula errors — the #1 source of wrong answers in this chapter. Zero extra time cost.
Read the question. Before writing anything, work through this sequence:
Step 1 — Identify the pool. How many total distinct items? (Check if items are from mixed categories — add them up first.)
Step 2 — Order test. Does swapping positions produce a different valid outcome? Yes → Permutation (nPr). No → Combination (nCr).
Step 3 — Restriction check. Is there a "must be together" or "must not be together" clause? If yes, use the complementary method: compute unrestricted total first, then subtract the forbidden count.
Step 4 — Compute. For nPr: multiply r terms downward from n. For nCr: same numerator, divide by r!. Use the symmetry swap if r > n/2.
Step 5 — Sanity check against options. If your answer is not among the choices, the most common error is mixing P with C, or forgetting the internal swap factor of 2 in the glue trick. Re-examine Step 2 first.
This five-step loop takes under 20 seconds to execute mentally once you have drilled it.
Why this question: Tests pure combination recognition — the most basic skill the exam checks. If you can't identify "selection from a group = nCr" instantly, this costs you.
Solving path: Pool = 9 fruits. Selecting 3 (a bag of fruit has no order) → Combination. C(9,3) = (9 × 8 × 7)/(3 × 2 × 1) = 504/6 = 84. Match option C.
Why this question: Tests permutation from a mixed pool. The trap is failing to add digits and letters before computing.
Solving path: Pool = 10 + 6 = 16 characters. Password = 4 characters, no repetition, order matters (a password is an arrangement) → P(16,4). Multiply 4 terms down: 16 × 15 × 14 × 13. Compute: 16 × 15 = 240, 240 × 14 = 3360, 3360 × 13 = 43680. Match option A.
Why this question: Tests whether you know the word "arrangements" signals permutation, and whether you can quickly count distinct letters in a word.
Solving path: BRIGHT — count the letters: B, R, I, G, H, T. Six distinct letters, no repeats. "3-letter arrangements" → P(6,3). Multiply 3 terms down: 6 × 5 × 4 = 120. Match option C.
Why this question: The classic restriction problem. Tests complementary counting — the most commonly tested advanced variation.
Solving path: Total arrangements of 6 people = 6! = 720. Arrangements where the two specific people are adjacent: glue them into 1 unit → 5 units, arranged in 5! = 120 ways, the glued pair can internally swap in 2! = 2 ways → 120 × 2 = 240. Valid = 720 - 240 = 480. Match option D.
Why this question: Sender-receiver problems look like combination problems but are permutations. This is a classic trap question.
Solving path: 8 members. Each sends a card to every other member = 7 cards per person. A sends to B and B sends to A are two separate cards. Total = 8 × 7 = 56. Match option B. (Do NOT compute C(8,2) = 28 — that double-counts by treating A→B and B→A as one event.)
Confusing sender-receiver with handshake problems. Handshakes between n people = C(n,2) because A-B handshake and B-A handshake are the same event. But greeting cards, phone calls (one person calls another), or "sends a message" — these are directional. Use n(n-1) = P(n,2). The word "send" almost always signals directionality.
Forgetting the internal swap in the glue trick. When two people are glued together, the unit can be arranged internally in 2! = 2 ways (AB or BA). Forgetting this factor of 2 gives you half the correct forbidden count and a wrong final answer.
Computing nCr when nPr is needed because the word "select" appears. Read the full question. "Select 3 people and assign them to three different posts" = permutation, not combination, even though "select" is used. The post-assignment signals order.
Applying formulas to repeated-letter words without adjustment. If a word has repeated letters (like LETTER with two T's and two E's), a 3-letter arrangement problem requires a different approach. The PYQs here use words with all distinct letters (BRIGHT). If you see repeated letters, the standard P(n,r) formula does not apply directly.
Wrong pool size in mixed-category problems. In the password question, students often compute P(10,4) using only digits or P(6,4) using only letters, forgetting to combine the pool. Always re-read what characters are available before plugging into the formula.
Using n! when only nPr is needed. "Arrange 3 out of 6 in a row" is P(6,3) = 120, not 6! = 720. The n! formula is only correct when ALL n items are arranged. The moment the problem says "arrange r out of n" with r < n, use P(n,r).